Use boundaries and graphs to find positive or negative outputs

Explain it like I am five

Picture a road crossing sea level. To find where it lies underwater, mark where it meets sea level. Those crossings separate the road into stretches. Then inspect a spot in each stretch to see whether the road is above or below. A boundary point plays that separating role in an inequality. A test point is a sample input in a stretch. A V has connected arms with no jumps, so it cannot move from above a comparison height to below it without crossing that height. The same method tells a shop which prices leave money after costs and which lose money. Positive output lies above zero. Negative output lies below zero.

The corner is (, 3); the graph is below zero outside − and .

Reminder

Negative fractions. − $1 over 2$ × 9 = − $9 over 2$ , and 3 = $6 over 2$ , so − $9 over 2$ + $6 over 2$ = − $3 over 2$ .
Flipping inequalities. Multiplying both sides of − $1 over 2$ |A| < −3 by −2 gives |A| > 6.
Set-builder notation. {x | x < 1} and {x : x < 1} both mean all real x such that x < 1. One divider is different from two absolute value bars.
Points on the axes. An x-intercept has output 0. Its input is the boundary number used in the interval answer.

Why it works. For a V and a horizontal line, their height difference has no jumps or breaks. This is called continuous. If that difference changed from positive to negative within an interval, it would have to pass through zero there. Solving their equality finds all such zero points first. With those points removed, a test point reveals the sign on each remaining interval. The graphical approach compares heights; the algebraic approach locates exact boundaries. A negative multiplier flips the comparison because it reverses number order. A downward V is negative outside its crossings because its arms descend away from its top.

RuleFind boundaries by solving the corresponding equality. Test one input inside each interval between boundaries, or compare graph heights. For f(x) < 0, choose the portions below the x-axis; for f(x) > 0, choose those above. Negative multiplication or division flips every comparison sign. Allow a boundary only when its original comparison permits equality.

The same idea, five ways

Say it

f of x is less than zero

Write it

Find the inputs whose outputs are negative, so their graph points lie below the x-axis.

In math

f(x) = − $1 over 2$ |4x − 5| + 3
f(x) < 0
x < − $1 over 4$ or x > $11 over 4$
(−∞, − $1 over 4$ ) ∪ ( $11 over 4$ , ∞)
{x | x < − $1 over 4$ or x > $11 over 4$ }
{x : x < − $1 over 4$ or x > $11 over 4$ }

Find the stretches of a road that lie below sea level.

See it

For the worked function, the distant arms fall below the horizontal axis.

The same idea, other ways

As sea level

Zero height is sea level. Above it means positive output; below it means negative output. The crossing inputs mark where the sign can change.

The graph is above zero between its zeros and below zero on its outer arms.

As a table of samples

Read a test input from the top row and its output directly underneath. A negative output selects that input's whole interval once every equality boundary has been found.

In the worked example, the first and third columns have negative outputs.

As comparing two heights

Hold the input fixed, then compare graph heights vertically. A V below the line y = 4 has an output smaller than 4. Its crossing inputs are where the heights match.

Interval for x	Test input x	Inside 4x − 5	Output f(x)	Passes f(x) < 0?
(−∞, − $1 over 4$ )	−1	−9	− $3 over 2$	Yes
(− $1 over 4$ , $11 over 4$ )	1	−1	$5 over 2$	No
( $11 over 4$ , ∞)	3	7	− $1 over 2$	Yes

.1Boundary and test-point method

Fenceposts divide a road into stretches. Solve the equality first to place every fencepost, then sample one input in each open stretch. The samples tell which stretches pass because these connected arms cannot cross the comparison height without another equality point.

For |x − 5| < 4, solve the equality |x − 5| = 4 to find boundaries 1 and 9.
The three open intervals are (−∞, 1), (1, 9), and (9, ∞).
The test inputs 0, 6, and 11 are shown in columns of the distance table.
For |x − 5| ≤ 4, the same interior interval works and its two boundaries are included.

The sample at 6 lies below height 4; the samples at 0 and 11 lie above it.

Reminder

Reading a table. An input is in the top row. Its output is directly beneath it in the same column.

Worked exampleRead the three interval tests

Use test inputs 0, 6, and 11 to solve |x − 5| < 4.

The sample at 6 lies below height 4; the samples at 0 and 11 lie above it.

Only the distance in the column under 6 is smaller than 4.

Only the open middle interval passes.

What it asks. Use one sample in each interval to find where the distance from five is smaller than four.

Plan. Find the equality boundaries, read the sample columns, then keep only intervals with distance below four.

Equality finds the boundaries; it does not decide which side passes.
A strict comparison excludes the boundary itself.

Set |x − 5| = 4 to find equality boundaries. Write x − 5 = −4 or x − 5 = 4.A comparison can switch sides only where the V reaches height 4, and inside values −4 and 4 reach that height.
Add 5 in each equation: x = 1 or x = 9. Check |1 − 5| = |9 − 5| = 4.This finds and confirms the two boundary inputs.
Test the left interval using x = 0: |0 − 5| = |−5| = 5. Read 5 in the column under 0.0 lies below 1, and 5 is greater than 4, so that interval fails.
Test the middle using x = 6: |6 − 5| = |1| = 1. Read 1 in the column under 6.6 lies between 1 and 9, and 1 is smaller than 4, so the middle interval passes.
Test the right interval using x = 11: |11 − 5| = |6| = 6. Read 6 in the column under 11.11 lies above 9, and 6 is greater than 4, so that interval fails.
Keep (1, 9) and exclude the boundaries.Only the middle test passes, and the boundary distance equals 4 rather than being smaller than 4.

Answer

(1, 9).

Check The algebraic chain −4 < x − 5 < 4 gives 1 < x < 9 too.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Use the equality answer {1, 9} as the strict inequality answer.

Equality locates the boundary numbers, and both of them fail the strict question. The inequality asks for an entire stretch of inputs.

✓ Instead: Test the intervals and choose (1, 9).

Tips and tricks

Pick test inputs with short arithmetic, but make sure each is inside its own interval.

.2Graphical comparison

Compare two road heights at the same horizontal position. The lower graph has the smaller output. Equality means the two heights meet, so inclusive comparisons keep their crossing inputs. A drawing shows the shape of the solution, while algebra supplies exact fractions when a screen is unclear.

Graphical approach: read relative heights.
Algebraic approach: solve the distance comparison exactly.
Use exact algebra for fractions that a sketch cannot locate precisely.

Inputs between the crossings make the V lower than the line.

Reminder

Input and output. At x = 6, the V has output 1 and the horizontal line has output 4, so 1 < 4.

Worked exampleAt or below a comparison height

Where is y = |x − 5| at or below y = 4?

Inputs between the crossings make the V lower than the line.

The inclusive comparison selects the closed interval [1, 9].

What it asks. Find where the V's height is no higher than the flat height four.

Plan. Find the crossing inputs and read where the V lies at or below the comparison line.

Both heights must be compared at the same input.
At or below includes the crossings.

Find crossings from |x − 5| = 4: x − 5 = −4 or x − 5 = 4, so adding 5 gives x = 1 or x = 9.This locates every input at which the two heights are equal.
Substitute the crossing inputs: |1 − 5| = |9 − 5| = 4.Both proposed boundaries really lie on the comparison line.
Read the picture between the crossings. At x = 5, |5 − 5| = 0 ≤ 4; the rising arms remain below the line until they reach 1 and 9.The V's lowest point lies in the middle and its two straight arms reach height 4 only at the boundaries.
Keep 1 ≤ x ≤ 9, giving [1, 9].At or below includes equality at both crossings.

Answer

[1, 9].

Check Subtracting the line height gives |x − 5| − 4 ≤ 0, the same inclusive inequality.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Compare the V at x = 1 with the line at a different input.

An inequality compares the two outputs for the same input, so mixing horizontal positions answers a different question.

✓ Instead: At each chosen x, compare the heights directly above that same x.

Tips and tricks

Use a graph for the regions and algebra for exact boundary fractions.

.3Positive and negative function outputs

Sea level separates above from below. For a graph, the horizontal axis has output zero. Positive output is above that axis and negative output is below it. A vertical reflection turns an upward V upside down, so its distant arms can be negative even when its middle is positive.

f(x) > 0 means graph above the x-axis.
f(x) < 0 means graph below the x-axis.
f(x) = 0 is a boundary for these strict sign questions.

The downward V is positive between its labeled zeros and negative outside them.

Reminder

Reflection. The outside minus in −|x| takes the opposite after the distance has been measured.

Worked exampleBoth signs of a reflected graph

Where is f(x) = −|x| + 2 positive, and where is it negative?

The downward V is positive between its labeled zeros and negative outside them.

The middle sample output is positive; both outside outputs are negative.

What it asks. Find the intervals where the reflected V has positive height and the intervals where it has negative height.

Plan. Find both zero inputs and test a sample in each of the three intervals.

A zero separates regions; it is not positive or negative itself.
Use zero as a middle sample and −3 and 3 as outside samples.

Set −|x| + 2 = 0 to find the crossings. Subtract 2: −|x| = −2. Multiply by −1: |x| = 2.A crossing has output zero. Undoing the outside operations exposes its exact distance.
The inside x equals −2 or 2. Check f(−2) = f(2) = −2 + 2 = 0.Both positions have distance 2, and substitution confirms both crossing inputs.
Test x = −3: f(−3) = −|−3| + 2 = −3 + 2 = −1. Read that negative output under −3.−3 is in the left interval and its output is below zero.
Test x = 0: f(0) = −|0| + 2 = 2. Read the positive output under 0.0 is in the middle interval and its output is above zero.
Test x = 3: f(3) = −|3| + 2 = −1. Read the negative output under 3.3 is in the right interval and its output is below zero.
Select the middle for positive output and the two outside intervals for negative output. Exclude both zeros in both answers.The sample signs select their intervals, and zero fails either strict sign question.

Answer

Positive: (−2, 2).
Negative: (−∞, −2) ∪ (2, ∞).

Check Rearranging f(x) > 0 gives |x| < 2; rearranging f(x) < 0 gives |x| > 2, matching the interval tests.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A negative input always gives a negative output.

Here f(−1) = −|−1| + 2 = 1 is positive even though the input is negative.

✓ Instead: Substitute into the function, then inspect the output sign.

Tips and tricks

Read positive and negative vertically, above and below the horizontal axis.

.4Prices that make a shop lose money

A shop can sell too cheaply or price an item so high that sales disappear. A model of its profit can rise toward one best price and fall on either side. Profit means money left after costs. A negative profit is a loss. In this model, finding negative outputs tells you which prices lose money.

P(x) = −2|x − 50| + 40 models a best profit of 40 at price 50.
P(x) < 0 asks for loss regions; it does not ask for negative prices.

The model loses money below price 30 or above price 70.

Reminder

Domain. The domain is the permitted input set. Here it is x ≥ 0, so a negative price is not allowed.

Worked exampleLoss on either side of the best price

A shop models profit by P(x) = −2|x − 50| + 40 for nonnegative prices x in dollars. Which prices make its profit negative?

Only permitted nonnegative prices in the two loss regions belong.

What it asks. Find the allowed prices at which the shop loses money.

Plan. Solve the negative output comparison, then keep only prices zero or more because the model restricts the inputs.

Subtract forty first.
Dividing by negative two flips the sign.
Keep only algebra answers also allowed by the stated nonnegative price domain.

Write −2|x − 50| + 40 < 0 and subtract 40: −2|x − 50| < −40.A negative profit is a loss, and subtraction removes the outside addition.
Divide by −2: |x − 50| > 20.Negative division reverses the sign and leaves the distance alone.
Write x − 50 < −20 or x − 50 > 20. Add 50: x < 30 or x > 70.Distance greater than 20 accepts either far side. Adding 50 finds the price limits.
Check the boundaries: P(30) = P(70) = −2 × 20 + 40 = 0.These prices cover costs exactly and do not make a negative profit.
Keep 0 ≤ x < 30 or x > 70, giving [0, 30) ∪ (70, ∞).The model permits only nonnegative prices, so drop the negative inputs from the first algebra interval.

Answer

0 ≤ x < 30 or x > 70
Price intervals: [0, 30) ∪ (70, ∞)

Check P(20) = −2 × 30 + 40 = −20 and P(80) = −20 confirm losses on both sides. P(50) = 40 confirms positive profit in the middle. P(0) = −100 + 40 = −60, so the allowed zero price is included.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Use (−∞, 30) ∪ (70, ∞) as the model's price answer.

The inequality accepts those algebra inputs, but the word problem explicitly restricts prices to zero or more.

✓ Instead: Keep [0, 30) ∪ (70, ∞).

Tips and tricks

After solving a word problem, keep the answer inside the input restrictions stated in the question.

Strategy: step by step

Decide which heights to compare, such as f(x) and 0, or |x − 5| and 4.
Set the two heights equal and solve. This finds every input at which the graph reaches the comparison height. Substitute those inputs back to confirm the crossings.
List the open intervals separated by those boundary inputs. Open means the boundary itself is not part of the interval being tested.
Pick one test point inside each interval and substitute it into the original comparison.
Keep every interval whose test point passes. Include a boundary only if substituting it into the original comparison passes too.
State the allowed inputs as an inequality, interval, or set-builder condition. If more than one interval works, use 'or' and ∪.

Strategy

Find positive or negative output regions

Is the comparison with zero?

YesSolve f(x) = 0 to find the inputs separating output sign regions.

NoSet the two graph heights equal to find their comparison boundaries.

↓

Are there any equality boundaries?

YesPut them in order and test each open interval they separate.

NoTest one input. For this continuous V and line comparison with no crossings, that same sign holds everywhere.

↓

Does a test point pass the original comparison?

YesKeep its interval.

NoLeave its interval out.

↓

Does the original sign allow equality?

YesTest and include each boundary whose equality passes.

NoExclude the boundaries, since equality fails a strict comparison.

Decide which heights to compare, such as f(x) and 0, or |x − 5| and 4.
Set the two heights equal and solve. This finds every input at which the graph reaches the comparison height. Substitute those inputs back to confirm the crossings.
List the open intervals separated by those boundary inputs. Open means the boundary itself is not part of the interval being tested.
Pick one test point inside each interval and substitute it into the original comparison.
Keep every interval whose test point passes. Include a boundary only if substituting it into the original comparison passes too.
State the allowed inputs as an inequality, interval, or set-builder condition. If more than one interval works, use 'or' and ∪.

Worked exampleA reflected function below zero

For f(x) = − $1 over 2$ |4x − 5| + 3, find where f(x) < 0.

The corner is (, 3); the graph is below zero outside − and .

Read the inside expression under each test input before taking its distance.

The output columns under −1 and 3 are negative; the column under 1 is positive.

Both outside rays belong, with exact boundaries − and excluded.

What it asks. Find every input whose output from this downward V is below zero.

Plan. Get the bars alone, find the equality crossings, test the three intervals, and exclude the zeros.

Subtract three before multiplying by negative two.
The negative multiplier flips the less than sign.
The first and third output columns are negative.

Subtract 3 from − $1 over 2$ |4x − 5| + 3 < 0: − $1 over 2$ |4x − 5| < −3.This undoes the outside vertical shift before comparing the distance.
Multiply both sides by −2: |4x − 5| > 6.−2 × (− $1 over 2$ ) = 1 and −2 × (−3) = 6. The negative multiplier flips the comparison.
To find the crossing boundaries, solve the equality |4x − 5| = 6.At distance 6 the original output is − $1 over 2$ × 6 + 3 = 0, exactly the comparison height.
Write 4x − 5 = −6 or 4x − 5 = 6.Only inside values −6 and 6 give distance 6.
Add 5: 4x = −1 or 4x = 11. Divide by 4: x = − $1 over 4$ or x = $11 over 4$ .These operations undo the inside shift and multiplier to find both crossing inputs.
Check the crossings: 4 × (− $1 over 4$ ) − 5 = −1 − 5 = −6, and 4 × $11 over 4$ − 5 = 11 − 5 = 6. Each gives f(x) = −3 + 3 = 0.Substitution confirms that both boundary inputs actually meet the horizontal axis.
Separate the inputs into (−∞, − $1 over 4$ ), (− $1 over 4$ , $11 over 4$ ), and ( $11 over 4$ , ∞).With every zero found, an output cannot change sign inside one of these intervals without another crossing.
Left test x = −1: 4(−1) − 5 = −9, so the inside table's column under −1 holds −9. Its distance is 9. Then f(−1) = − $1 over 2$ × 9 + 3 = − $9 over 2$ + $6 over 2$ = − $3 over 2$ .Multiplying by negative one half gives − $9 over 2$ , and writing 3 as $6 over 2$ lets the halves subtract. The negative result selects the left interval.
Middle test x = 1: 4 × 1 − 5 = −1. Then f(1) = − $1 over 2$ × 1 + 3 = − $1 over 2$ + $6 over 2$ = $5 over 2$ . Read the positive output in the column under 1.1 is between the exact boundaries, and its positive output rejects that entire middle interval.
Right test x = 3: 4 × 3 − 5 = 7. Then f(3) = − $7 over 2$ + $6 over 2$ = − $1 over 2$ . Read the negative output in the column under 3.3 is beyond the right boundary, and the negative result selects the right interval.
Find the corner by setting 4x − 5 = 0: add 5 and divide by 4 to obtain x = $5 over 4$ . Substitute: f( $5 over 4$ ) = − $1 over 2$ |0| + 3 = 3.Zero inside identifies the V's turning input. Substitution gives its height and explains the graph's corner ( $5 over 4$ , 3).
Moving one unit farther from the corner changes |4x − 5| by 4. The height change is − $1 over 2$ × 4 = −2.Factoring 4x − 5 = 4(x − $5 over 4$ ) gives the same function as −2|x − $5 over 4$ | + 3, so the picture's arms descend two units per step away.
Keep the two outside intervals and leave their boundary inputs out.The outside tests are negative, while each boundary output equals zero and fails the strict question.

Answer

x < − $1 over 4$ or x > $11 over 4$
Interval: (−∞, − $1 over 4$ ) ∪ ( $11 over 4$ , ∞)
Set-builder: {x : x < − $1 over 4$ or x > $11 over 4$ }

Check The between method checks the rejected middle: f(x) ≥ 0 gives |4x − 5| ≤ 6, hence −6 ≤ 4x − 5 ≤ 6. Add 5 and divide by 4 to get − ≤ x ≤ . Every input outside that closed interval has negative output, confirming the selected two rays.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: parent graph above zero

Where is f(x) = |x| positive?

All real inputs except zero give positive distance.

What it asks. Find where the basic distance graph is above zero.

Plan. Find its only zero and exclude that one input.

Every input other than zero has positive distance.
The word positive excludes output zero.

|x| = 0 gives x = 0.Setting the output to zero finds the only point separating the two positive stretches. Zero distance occurs only at x = 0; substitution gives |0| = 0.
Every x < 0 and every x > 0 gives positive distance.Every position other than zero has positive distance from the origin.

Answer

(−∞, 0) ∪ (0, ∞).

Check f(−1) = f(1) = 1, while f(0) = 0 and is excluded.

Rung 2Rung 2: upward V below zero

Where is f(x) = |x − 2| − 3 negative?

The upward V lies below zero between −1 and 5.

Only the middle sample has negative output.

Both crossing inputs are excluded from the negative region.

What it asks. Find where the upward V lies below the horizontal axis.

Plan. Find its two crossings and test the middle and both outside intervals.

Set output zero to find boundary inputs.
The center lies in the middle interval.

Set |x − 2| − 3 = 0 to find the crossings. Add 3: |x − 2| = 3.Output zero marks the boundaries where the sign can change.
Write x − 2 = −3 or x − 2 = 3. Add 2: x = −1 or x = 5.These are the only inside values at distance 3, so solving finds both boundary inputs.
Check f(−1) = |−3| − 3 = 0 and f(5) = |3| − 3 = 0.Both proposed boundaries really lie on the horizontal axis.
Read the test columns: f(−2) = |−4| − 3 = 1, f(2) = |0| − 3 = −3, and f(6) = |4| − 3 = 1.The table shows a positive left sample, a negative middle sample, and a positive right sample.
Keep (−1, 5).Only the middle interval has negative output, and its zero endpoints do not pass the strict comparison.

Answer

(−1, 5).

Check Algebra gives |x − 2| < 3, hence −1 < x < 5.

Rung 3Rung 3: downward V above zero

Where is f(x) = −3|x + 2| + 6 positive?

The downward V is positive between its two zeros −4 and 0.

The strict positive region excludes both zero inputs.

What it asks. Find inputs making the downward V above zero.

Plan. Isolate the distance, flip after negative division, and solve the resulting between chain.

Undo the plus six first.
Negative three outside the bars reverses the comparison.

Subtract 6 from −3|x + 2| + 6 > 0: −3|x + 2| > −6.This removes the outside shift before inspecting the distance.
Divide by −3: |x + 2| < 2.The negative divisor flips the greater than sign to a less than sign.
Write −2 < x + 2 < 2.Distance less than two requires both inside limits.
Subtract 2 everywhere: −4 < x < 0.This removes the inside addition and identifies every input giving positive output.

Answer

(−4, 0).

Check At the vertex −2 the output is 6; at the boundaries −4 and 0 it is zero, so only the middle is positive.

Rung 4Rung 4: a fractional reflected graph including its zeros

Where is f(x) = − $1 over 2$ |2x − 6| + 3 at least zero?

Factoring 2x − 6 gives −|x − 3| + 3, a downward V at or above zero on [0, 6].

Filled dots include the zero outputs at x = 0 and x = 6.

What it asks. Find every input making the output zero or positive.

Plan. Undo the outside shift and negative half, solve between, and include the two crossing inputs.

At least zero allows equality.
Multiplying by negative two flips the sign.
Keep both ends after solving the between chain.

Write − $1 over 2$ |2x − 6| + 3 ≥ 0 and subtract 3: − $1 over 2$ |2x − 6| ≥ −3.The question compares the entire output with zero, and subtraction removes the shift.
Multiply by −2: |2x − 6| ≤ 6.The multiplier cancels the negative half and reverses the sign, while retaining equality.
Write −6 ≤ 2x − 6 ≤ 6.Distance at most six gives a between chain with both boundaries included.
Add 6 everywhere: 0 ≤ 2x ≤ 12.This removes the inside subtraction from all comparisons.
Divide every part by 2: 0 ≤ x ≤ 6.The positive divisor isolates the inputs whose outputs are at least zero.
Write [0, 6].Both endpoint outputs are zero and therefore pass the stated inclusive comparison.

Answer

0 ≤ x ≤ 6
Interval: [0, 6]

Check f(0) = − × 6 + 3 = 0, f(6) = 0, and f(3) = 3, so the endpoints and middle pass. f(−1) = − × 8 + 3 = −1 and f(7) = −1 fail outside.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Multiply −|4x − 5| < −3 by −2 and keep the less than sign.

The negative multiplier reverses order. Keeping the sign chooses the middle, where f(1) = is positive and fails the original question.

✓ Instead: Use |4x − 5| > 6, then select the outside intervals.

✗ Not this: Test only one input before finding all equality boundaries.

A point in the middle of a downward V can have positive output while points on either far arm have negative output.

✓ Instead: Find every boundary first, then test one point in each remaining interval.

✗ Not this: Include zeros when the question asks for negative output.

At a zero, the output equals 0 and is not smaller than 0.

✓ Instead: Exclude zeros for a strict negative question; include them when the question asks for output at most zero.

Tips and tricks

One test point per interval, after every crossing is found.
Check the test point in the original formula. That catches an incorrect sign flip made while isolating.
Negative output means below the x-axis. It does not mean a negative input.

Trap. Multiplying by −2 and keeping the less than sign would select the center, where this function is positive. Flip the sign immediately and confirm your answer with an actual output from each chosen interval.